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<h1>Dome Height</h1>

<p>The point P5 in figure 45 represents the bridge position, measured from the top edge of the body. The dome height at this position is measured down perpendicular to the point P6, where the vertical line crosses the circumference of the dome circle. </p>

<p class="imgbox"><a name="fig-058"></a><img src="../img/fig-058.png" style="max-width: 400px;"/><br />Figure 58: Calculating the dome height at bridge position.</p>

<h2>What We Know</h2>

<pre class="formula">
(1)	P0P1 = P0P2 = P0P3 = DR = dome radius
(2)	P1P2 = BL = body length
(3)	P1P5 = BP = bridge position
</pre>

<h2>What Is Unknown</h2>

<pre class="formula">
(4)	P5P6 = DH = dome height at bridge position
</pre>

<h2>Calculation Path</h2>

<p>The line P0-P3 is the bisector of the triangle P1-P0-P2, and divides the body length P1-P2 into halves at P4. </p>

<pre class="formula">
(5)	P1P4 = P4P2 = P1P2 / 2
</pre>

<p>P3 is also the point of maximum dome height P4P3.</p>

<p>The triangle P0-P4-P1 has a square angle at P4, and P1-P4 is know from equation (5), so we can calculate the side P0-P4 by applying the pythagorean theorem:</p>

<pre class="formula">
(6)	P0P1<sup>2</sup> = P1P4<sup>2</sup> + P4P0<sup>2</sup>
(7)	P0P4<sup>2</sup> = P0P1<sup>2</sup> &minus;  P1P4<sup>2</sup>
(8)	P0P4 = sqrt( P0P1<sup>2</sup> &minus; P1P4<sup>2</sup> )
</pre>

<p>Adding a horizontal helper through P6 parallel to the reference line P1-P2 creates a rectangle P4-P5-P6-P7:</p>

<p class="imgbox"><a name="fig-057"></a><img src="../img/fig-057.png" style="max-height: 200px;"/><br />Figure 57</p>

<p>Because opposite lines are parallel, they also have the same length:</p>

<pre class="formula">
(9)	P4P5 = P7P6
(10)	P5P6 = P4P7
</pre>

<p>The length P4-P5 can be calculated from body length and bridge position:</p>

<pre class="formula">
(11)	P1P5 = P1P4 + P4P5
(12)	P4P5 = P1P5 - P1P4
</pre>

<p>Substitution of P1P4 with equation (5) gives us</p>

<pre class="formula">
(13)	P4P5 = P1P5 - P1P2 / 2
</pre>

<p>The helper line givs us also the triangle P0-P7-P6, with a square angle at P7. Equations (10) and (13) tell us that P6-P7 is known, so we can calculate the side P0P7:</p>

<pre class="formula">
(14)	P0P7<sup>2</sup> + P7P6<sup>2</sup> = P6P0<sup>2</sup>
(15)	P0P7<sup>2</sup> = P6P0<sup>2</sup> &minus;  P7P6<sup>2</sup>
(16)	P0P7 = sqrt( P6P0<sup>2</sup> &minus;  P7P6<sup>2</sup> )
</pre>

<p>We also know that P4 divides P0-P7:</p>

<pre class="formula">
(17)	P0P7 = P0P4 + P4P7
</pre>

<p>P0-P4 is known from equation (8), and P0-P7 is known from equation (16), which allows us to determine P4-P7: </p>

<pre class="formula">
(18)	P4P7 = P0P7 &minus; P0P4 
</pre>

<p>Introduction of equation (10) gives us the the value for the desired dome height at bride position P5:</p>

<pre class="formula intermediate">
(19)	P5P6 = P0P7 &minus; P0P4 
</pre>

<h2>Replacment of Intermediate Terms</h2>

<p>We could stop here because we are able to calculate the desired value, but in order to get a single formula with only original values we have to replace all intermediate terms. This makes the formula appear much more complicated, but allows us to use the complete formula in one piece elsewhere (a spreadsheet calculation program for instance) without need for intermediate calculation steps and associated variables.</p>

<p>Substituion of P0-P4 in (19) with equation (9):</p>

<pre class="formula">
(20)	P5P6 = P0P7 &minus; sqrt( P0P1<sup>2</sup> &minus; P1P4<sup>2</sup> )
</pre>

<p>Substitution of P1-P4 with equation (5):</p>

<pre class="formula">
(21)	P5P6 = P0P7 &minus; sqrt( P0P1<sup>2</sup> &minus; (P1P2 / 2)<sup>2</sup> )
</pre>

<p>Substitution of P0-P7 with equation (17):</p>

<pre class="formula">
(22)	P5P6 = sqrt( P0P6<sup>2</sup> &minus;  P6P7<sup>2</sup> ) &minus; sqrt( P0P1<sup>2</sup> &minus; (P1P2 / 2)<sup>2</sup> )
</pre>

<p>Substitution of P6P7 with equations (9) and (12):</p>

<pre class="formula">
(23)	P5P6 = sqrt( P0P6<sup>2</sup> &minus;  (P1P5 &minus; P1P4)<sup>2</sup> ) &minus; sqrt( P0P1<sup>2</sup> &minus; (P1P2 / 2)<sup>2</sup> )
</pre>

<p>Substitution of P1-P4 with equation (5):</p>

<pre class="formula">
(24)	P5P6 = sqrt( P0P6<sup>2</sup> &minus;  (P1P5 &minus; P1P2 / 2)<sup>2</sup> ) &minus; sqrt( P0P1<sup>2</sup> &minus; (P1P2 / 2)<sup>2</sup> )
</pre>

<p>Now we can replace the generic names with familiar names according to equations (1) to (4) to find the final formula:</p>

<pre class="formula final">
(25)	DH = sqrt( DR<sup>2</sup> &minus;  (BP &minus; BL / 2)<sup>2</sup> ) &minus; sqrt( DR<sup>2</sup> &minus; (BL / 2)<sup>2</sup> )
</pre>



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